Question

Interference fringes were produced in Youngs double slit experiment using light of wavelength $5000\mathring{A}$. When a film of material $2.5\times 10^{-3}cm$ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to $20$ fringe widths. The refractive index of the material of the film is

• A. $1.25$
• B. $1.33$
• C. $1.4$
• D. $1.5$

Answer 1

Answer: (C)

Fringe width, $\beta =\frac{\lambda D}{d}...(i)$
where $D$ is the distance between the screen and slit and $d$ is the distance between two slits.
When a film of thickness $t$ and refractive index $\mu$ is placed over one of the slit, the fringe pattern is shifted by distance $S$ and is given by
$S=\frac{(\mu -1)tD}{d}...(i)$
Given : $S=20\beta ...(iii)$
From equations $(i),(ii)$ and $(iii)$ we get,
$(\mu -1)t=20\lambda$
or $(\mu -1)=\frac{20\lambda }{t}$
$=\frac{20\times 5000\times 10^{-8}cm}{2.5\times 10^{-3}cm}$
$\mu -1=0.4$ or $\mu =1.4$